Fix an epistemic Kripke model \(\mathfrak {M}=(S,E,F)\)  with a possible world \(s\in S\) , and suppose its corresponding generated argument Kripke model is \(\mathfrak {N}_g=(\mathbb {W},A,G)\) , where we have \(\lbrace s\rbrace \in \mathbb {W}\) . Assume \(\mathfrak {N}_g\)  is nontrivial. Thus once again, with respect to this fixed argument Kripke model \(\mathfrak {N}_g\)  and this fixed possible argument \(\lbrace s\rbrace \) , suppose its corresponding generated epistemic Kripke model is \(\mathfrak {M}_{gg}=(S^{\prime },E^{\prime },F^{\prime })\) . We reasonably expect that \(\mathfrak {M}_{gg}\) , even though may not be exactly identical to the original \(\mathfrak {M}\) , must resemble \(\mathfrak {M}\)  to a certain extent. Nonetheless, in order to pin down such intuition precisely, we firstly have to determine which possible world in \(S^{\prime }\)  should correspond to the original \(s\in S\) , namely, represent the currently real world.

For any \(t\in S\)  namely \(\lbrace t\rbrace \in \mathbb {W}\) , there does not exist \(U\in \mathbb {W}\)  such that \(U\le \lbrace t\rbrace \wedge \lbrace t\rbrace \nleq U\) , where relation \(\le \)  is defined in Definition .


To a contradiction suppose such \(U\)  exists, since \(\lbrace t\rbrace \nleq U\) , there exists \(p\in P\) , \(V\in \mathbb {W}\)  such that \((V,U)\in A(p)\wedge (V,\lbrace t\rbrace )\notin A(p)\) , hence there does not exist \(v\in V\)  such that \((v\in F(p)\wedge t\notin F(p))\vee (v\notin F(p)\wedge t\in F(p))\) , namely for any \(v\in V\) , \(v\in F(p)\iff t\in F(p)\) . And then as \((V,U)\in A(p)\) , it is easy to see that for any \(u\in U\) , \(v\in V\) , \(u\in F(p)\iff v\notin F(p)\iff t\notin F(p)\) , thus \((U,\lbrace t\rbrace )\in A(p)\)  but of course \((U,U)\notin A(p)\) , therefore \(U\nleq \lbrace t\rbrace \) , a contradiction.


For any \(t\in S\) , \(\tau (t)=\lbrace U\in \mathbb {W}\mid \lbrace t\rbrace \le U\rbrace \subseteq \mathbb {W}\)  is an ultrafilter.


Since \(\lbrace t\rbrace \le \lbrace t\rbrace \) , \(\lbrace t\rbrace \in \tau (t)\)  so \(\tau (t)\ne \emptyset \) . For any \(U\in \tau (t)\) , any \(W\in \mathbb {W}\) , if \(U\le W\)  then by transitivity \(\lbrace t\rbrace \le W\) , hence \(W\in \tau (t)\) . For any \(U,V\in \tau (t)\) , \(\exists \lbrace t\rbrace \in \tau (t)\)  such that \(\lbrace t\rbrace \le U\wedge \lbrace t\rbrace \le V\) . Thus \(\tau (t)\)  is a filter. To a contradiction suppose \(\tau (t)=\mathbb {W}\) , then any proper filter \(\mathbb {U}\subset \mathbb {W}\)  is not an ultrafilter, namely there are no ultrafilters, contradicting our assumption that \(\mathfrak {N}_g\)  is nontrivial. Thus \(\tau (t)\subset \mathbb {W}\)  is a proper filter. To a contradiction suppose there exists \(\mathbb {U}\subseteq \mathbb {W}\)  such that \(\tau (t)\subset \mathbb {U}\)  and that \(\mathbb {U}\)  is a filter, then there exists \(U\in \mathbb {U}\setminus \tau (t)\) , as \(\mathbb {U}\)  is a filter, there exists \(V\in \mathbb {U}\)  such that \(V\le U\wedge V\le \lbrace t\rbrace \) , and because \(U\notin \tau (t)\) , we also have \(V\notin \tau (t)\) , namely \(\lbrace t\rbrace \nleq V\) , contradicting the previous Lemma . Thus \(\tau (t)\)  is an ultrafilter.

[Principal Ultrafilter]
As Proposition  shows, for any \(t\in S\) , we call \(\tau (t)\in S^{\prime }\)  as \(t\) 's principal ultrafilter.

Although the definition here is slightly different from ordinary principal ultrafilters over a Boolean algebra [1]}, readers should feel convinced that the core idea is the same and so we are quite justified for usage of the same name. Then of course, the current possible world in the “generated generated” epistemic Kripke model \(\mathfrak {M}_{gg}\)  should be \(\tau (s)\in S^{\prime }\) , namely, the principal ultrafilter of the original current world \(s\in S\) .
